To show that it is continuous I used polar coordinates to show that the limit at the origin is indeed 0 and so it must be continuous everywhere since the limit obviously equals the function everywhere else
To determine if it is differentiable, I found the partial derivative with respect to x and said that it does not exist at (0,0) therefore the function is not differentiable at the origin.
Thanks
Best Answer
It's easy to prove that $\lim_{(x,y)\to (0,0)}f(x,y)=0=f(0,0)$ (see this) hence $f(x,y)$ is continous at $(0,0)$. To determine if $f(x,y)$ is differentiable, we need the values of the derivatives $f_{x}(0,0)$ and $f_{y}(0,0)$. By definition of partial derivatives:
$\\ f_{x}(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}0=0 \\ \\ f_{y}(0,0)=\lim_{k\to 0}\frac{f(0,k)-f(0,0)}{k}=\lim_{k\to 0}0=0$
Now $f(x,y)$ is differentiable in $(0,0)$ if and only if $$\lim_{(h, k)\to (0,0)}\frac{f(h,k)-f(0,0)-f_{x}(0,0)k-f_{y}(0,0)k}{\sqrt{h^2+k^2}}=0$$
which becomes
$$\lim_{(h, k)\to (0,0)}\frac{h k^3}{(h^2+k^4)\sqrt{h^2+k^2}}=0$$
but if you take $h=k^2$, the restriction is $$\lim_{k\to 0}\frac{k}{2\sqrt{k^2+k^4}}=\lim_{k\to 0}\frac{k}{2|k|\sqrt{1+k^2}}=\begin{cases}\frac{1}{2}&\mbox{if} \ k\to 0^{+}\\ -\frac{1}{2}&\mbox{if}\ k\to 0^{-}\end{cases}$$
On the path $(k^2,k)$ the limit $$\lim_{(h, k)\to (0,0)}\frac{h k^3}{(h^2+k^4)\sqrt{h^2+k^2}}\ne 0$$ hence $f(x,y)$ is not differentiable in $(0,0)$.